The length of a rectangular garden is 8 feet longer than the width. The garden is surrounded by a 4-foot sidewalk. The sidewalk has an area of 320 square feet. Find the dimensions of the garden.

1. The information given in the problem is:

- The length of a rectangular garden is 8 feet longer than the width.

- The garden is surrounded by a 4-foot sidewalk.

- The area of the sidewalk is 320 ft ².

2. So, the length of the rectangular garden is:

L1=8+W1

3. The formula for calculate the area of the sidewalk, is:

A2=L2xW2

"A2" is the area of the sidewalk (A2=320 ft²).

"L2" is the length of the sidewalk.

"W2" is the widht of the sidewalk.

4. The length of the sidewalk (L2) is:

L2=L1+4+4 (4 feet on each side)

L2=L1+8

5. When you substitute L1=8+W1 into the equation L2=L1+8, you obtain:

L2=8+W1+8

L2=W1+16

6. The widht of the sidewalk is:

W2=W1+4+4

W2=W1+8

7. Now, you must substitute the length and the widht of the sidewalk into the formula A2=L2xW2:

A2=L2xW2

A2 = (W1+16) (W1+8)

320=W1²+16W1+8W1+128

W1²+16W1+8W1+128-320=0

W1²+16W1+8W1-192=0

8. When you solve the quadratic equation, you obtain the value of W1:

W1=16.97 ft

9. Finally, you must substitute the value of W1 into the formula L1=8+W1:

L1=8+W1

L1=8+16.97

L1=24.97 ft

10. Therefore, the dimensions of the garden are:

L1=24.97 ft

W1=16.97 ft